/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES
We consider the system theBenchmark.

We are asked to determine termination of the following first-order TRS.

  f : [o * o] --> o 
  h : [o] --> o 
  minus : [o] --> o 

  minus(minus(X)) => X 
  minus(h(X)) => h(minus(X)) 
  minus(f(X, Y)) => f(minus(Y), minus(X)) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  minus(minus(X)) >? X 
  minus(h(X)) >? h(minus(X)) 
  minus(f(X, Y)) >? f(minus(Y), minus(X)) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  f = \y0y1.y0 + y1 
  h = \y0.3 + y0 
  minus = \y0.3y0 

Using this interpretation, the requirements translate to:

  [[minus(minus(_x0))]] = 9x0 >= x0 = [[_x0]] 
  [[minus(h(_x0))]] = 9 + 3x0 > 3 + 3x0 = [[h(minus(_x0))]] 
  [[minus(f(_x0, _x1))]] = 3x0 + 3x1 >= 3x0 + 3x1 = [[f(minus(_x1), minus(_x0))]] 

We can thus remove the following rules:

  minus(h(X)) => h(minus(X)) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  minus(minus(X)) >? X 
  minus(f(X, Y)) >? f(minus(Y), minus(X)) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  f = \y0y1.1 + y0 + y1 
  minus = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[minus(minus(_x0))]] = 4x0 >= x0 = [[_x0]] 
  [[minus(f(_x0, _x1))]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[f(minus(_x1), minus(_x0))]] 

We can thus remove the following rules:

  minus(f(X, Y)) => f(minus(Y), minus(X)) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  minus(minus(X)) >? X 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  minus = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[minus(minus(_x0))]] = 6 + x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  minus(minus(X)) => X 

All rules were succesfully removed.  Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.